Python lxml 使用
lxml,是python中用来处理xml和html的功能最丰富和易用的库
from lxml import etree from lxml import html h = ''' <html> <head> <meta name="content-type" content="text/html; charset=utf-8" /> <title>友情链接查询 - 站长工具</title> <!-- uRj0Ak8VLEPhjWhg3m9z4EjXJwc --> <meta name="Keywords" content="友情链接查询" /> <meta name="Description" content="友情链接查询" /> </head> <body> <h1 class="heading">Top News</h1> <p style="font-size: 200%">World News only on this page</p> Ah, and here's some more text, by the way. <p>... and this is a parsed fragment ...</p> <a href="http://www.cydf.org.cn/" rel="nofollow" target="_blank">青少年发展基金会</a> <a href="http://www.4399.com/flash/32979.htm" target="_blank">洛克王国</a> <a href="http://www.4399.com/flash/35538.htm" target="_blank">奥拉星</a> <a href="http://game.3533.com/game/" target="_blank">手机游戏</a> <a href="http://game.3533.com/tupian/" target="_blank">手机壁纸</a> <a href="http://www.4399.com/" target="_blank">4399小游戏</a> <a href="http://www.91wan.com/" target="_blank">91wan游戏</a> </body> </html> ''' # 第一种使用方法 page = etree.HTML(h) #hrefs = page.xpath('//a') href = page.cssselect('a') for href in hrefs: print(href.attrib) 第二种使用方法 def parse_from(): tree = html.fromstring(h) for href in tree.cssselect('a'): #for hfre in tree.xpath('//a'): a = href print(a.text) print(a.attrib) paese_from() parse_from()